Lithium oxide is used to absorb carbon dioxide:

Li2O(s) + CO2(g) → Li2CO3(s)

What volume of CO2 can 6.77 g of Li2O absorb if the CO2 pressure is 3.5 × 10−4 atm and the temperature is 295 K?

Water is a polar molecule because in the O-H bond of water, Oxygen is more electronegative due to which partial positive charge and partial negative charge is created on hydrogen and oxygen respectively, resulting in a dipole moment.

A dipole moment is created between two atoms having a suitable electronegativity (ability to pull electrons more towards itself) difference.

A difference in electronegativity between two atoms creates a partial positive charge and a partial negative charge on the atoms. The electrons are pulled towards the atom having partial negative charge. The resultant direction of the atoms in which they get pulled gives the dipole moment of the bond.

If the resultant dipole is zero then the bond is considered to be non-polar

water is an example of a polar molecule as in the bent structure of the molecule, with oxygen at the center, surrounded by two H atoms on both sides, the electrons of the two O-H bonds are pulled towards O, resulting in a dipole moment in the upward direction. Thus, making water a polar molecule.

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The ΔH of the reaction for the formation of graphite from diamond is equal to -1.9 KJ/mol.

The heat of formation can be described as the amount of heat absorbed or evolved when 1 mole of a substance is formed from its constituent elements, each substance being in its normal physical state.

We have to rearrange the provided equations in a way in which all molecules of O₂ and CO₂ can be eliminated:

2C (diamond) + 2O₂ (g) → 2CO₂ (g)     ΔH₁= 2 × (-395.4 KJ)

2 CO₂ (g) → 2 CO(g) + O₂(g)            ΔH₂ = 566.0 KJ

CO₂(g) → C (graphite) + O₂(g)          ΔH₃= - 1 × (-393.5 KJ)

2CO (g) → C (graphite) + CO₂(g)      ΔH₄= -172.5 KJ

When we eliminate the molecules that appear both in reactants and products, the total chemical reaction is the following:

2C (diamond) → 2C (graphite)

ΔH (total) = ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄

ΔH (total) = 2 × (-395.4 KJ) + 566.0 KJ + (-1 × (-393.5 KJ)) - 172.5 KJ

ΔH (total) = 347.2 KJ

This is for 2 mol of C (diamond) which is converted into 2 mol of C (graphite). To find ΔH for the reaction of 1 mol C (diamond) to 1 mol (graphite) we have to divide it into 2:

ΔH = - 3.8/2 = -1.9 KJ/mol

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A well-known NMR technique called diffusion ordered spectroscopy (DOSY) provides diffusion coefficients for specific resonances in NMR spectra.

The main applications of DOSY are the analysis of small molecule mixtures and the oligomeric state of biomolecules. The spreading Self-Diffusion (SD)-NMR and diffusion ordered spectroscopy are two common names for the NMR technology (DOSY). This is accomplished by fusing magnetic field gradients, which encode spatial information, with radio-frequency pulses, which are commonly employed in NMR spectroscopy. Magnetic field gradients are used by Diffusion Ordered Spectroscopy (DOSY) to examine diffusion processes in both solid and liquid materials. Liquid structure is discovered using NMR spectroscopy. It is also used to figure out how soluble chemical molecules are structured.

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The reaction is given :    ²³⁴Pa₉₁  ----->   X  +   ⁰e₋₁ , the equation is balanced and the nucleus represented by X is ²³⁴U₉₂.

The balanced reaction is given as :

²³⁴Pa₉₁  ----->   X  +   ⁰e₋₁

This reaction is the balanced reaction and X is represented the nucleus and is given as :  ²³⁴U₉₂.

The given reaction is the nuclear reaction. the nuclear reaction is the reaction in which the nuclides are produced by the collision between the two atoms nuclei. the nuclear reaction are of the two types the nuclear fission and the nuclear fusion reaction. the nuclear fission is the splitting of the large nuclei in to the smaller ones.

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Acid concentration: 0.49 M KOH concentration is 0.49 MKa1=1.0 x 104 eq. rm

a remedy for Take the example of titrating 100 mL of 0.100 M HOCl with 0.100 M KOH at 25 °C. HOCl has a Ka of 3.5 10-8. Calculate the pH after 0.0 mL of KOH in Part 1.

Consider titrating 0.115 M RbOH in a 25.0 mL sample with 0.100 M HCl. Figure out each quantity. The initial pH and the amount of acid that must be added to achieve the equivalence point are also factors. c. the pH after adding 5.0 mL of acid. pH at the equivalency point (d). e. the pH following the addition of 5.0 mL of acid over the equivalency point.

Think about titrating 100 mL of 0.59 M H 3 a 1-3 a2-6 a3. 1) Determine the pH following

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The Solubility Of Fe(OH)₂ In Water At 25 °C Is Measured To Be 5.2x 10⁻⁴ Fe(OH)₂ Ksp value is 5.6 × 10⁻¹⁰.

The chemical reaction is given as follows :

Fe(OH)₂ ⇄  Fe²⁺    +   2OH⁻

Ksp expression for the given reaction is given as follows :

Ksp = [Fe²⁺] [OH⁻]²

the solubilty S =  5.2x 10⁻⁴

so , the Ksp = S × (2S)²

                    = 4S³

Ksp = 4 ×S³

      = 4 × (  5.2x 10⁻⁴ )³

      = 4 × 140.6 × 10⁻¹²

      = 5.6 × 10⁻¹⁰

The solubility product value for Fe(OH)₂   that is Ksp value is 5.6 × 10⁻¹⁰.

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The diagrams of zonation in lakes and oceans with correct targets are :

a) fresh plant may be rooted here

b) littoral zone

c) limnetic zone

d) alternately dry and submerged

e) extends to edge of the continental shelf

f) generally low nutrient levels

g) light penetrates here

h) no photosynthesis occurs here

i) benthic zone

There are zones which divided the lakes. they are separated by the water from the top to bottom and the side to the side. that is named as : the littoral zone , the limnetic zone, the profundal zone , the benthic zone and the euphotic zone . the waters temperature will be affected the density. there are several zones in the ponds and the lakes. these are dived by the water column.

The limnology divides the lakes in the three types of zones ., that is the limnetic zone, the littoral zone and the benthic zone.

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Since CH3CH2CH2NH2 may form intermolecular hydrogen bonds, it has a higher boiling point than other compounds. The hydrogen atoms in (CH3)3N (C H 3) 3 N are joined to the carbon atoms.

Since carbon is not a particularly electronegative atom, it cannot donate hydrogen. There aren't any hydrogen atoms to accept, despite the fact that nitrogen is a strong hydrogen atoms. As a result, CH3CH2CH2CH3 is a nonpolar molecular complex in which induced dipole forces are the only intermolecular forces. When both the donor atom, D, and the acceptor atom, A, are among the highly electronegative elements O, N, or F, hydrogen bonding occurs, as shown by the symbol D-H—-A.

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Label for each of the compounds given in the question are: 1) Isomer, 2) Isomer, 3) Resonance Structure, 4) Isomer 5) Neither.

A group of two or more Lewis structures known as resonance structures describe the electronic bonding of a single polyatomic species, including fractional bonds and fractional charges. The number of electrons in resonance structures should be constant; do not add or subtract any electrons. (Count the electrons to determine their number). Every resonance structure complies with the standards for writing Lewis Structures. Structure hybridization must be constant. By cyclically moving elections, it is possible to depict the two resonance configurations of the benzene ring. Multiple resonance structures can be found in conjugated double bonds.

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The chemical species in solution S2 Fe(s)+NiCl2(aq) → FeCl2(aq)+Ni(s)

Fe(s) ------> [tex]Fe^{2+}[/tex] + 2e                              (Oxidation)

[tex]Ni^{2+}[/tex] (aq) + 2e ------> Ni(s)                        (Reduction)

Galvanic Cell

Galvanic cells employ the electrical energy generated by the transport of electrons during a redox reaction to carry out practical electrical work. By separating the oxidation and reduction half-reactions and connecting them with a wire so that the electrons must pass through the wire, it is possible to gather the electron flow.

Write a balanced equation for the half-reaction that happens at the cathode of this cell = Ni (aq) + 2e ------> Ni(s)

Write a balanced equation for the half-reaction that happens at the anode of this cell = Fe(s) ------> [tex]Fe^{2+}[/tex] + 2e.

Of what substance is E1 made? = Fe

of what substance is E2 made? = Ni

What are the chemical species in solution S1? = [tex]Fe^{2+}[/tex]

What are the chemical species in solution S2? = [tex]Ni^{2+}[/tex]

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As per the given balanced reaction, one mole or 63.5 g of Cu produces 2 moles or 215.6 g of silver metal. Hence, 5 g of copper produces16.9 g of Ag.

A displacement reaction is a type of reaction, in which one element or group of the reactant is displaced by the attacking group or element. The reaction between copper and silver nitrate is an example of single displacement reaction.

Given the balanced equation  of the reaction. One mole of Cu produces 2 moles of Ag.

atomic mass of Cu = 63.5 g/mol

atomic mass of Ag = 107.8 g/mol

mass 2 moles of Ag = 2 × 107.8 =215.6 g.

Hence, 63.5 g of Cu produces 215.6 g of Ag.

The mass of Ag produced by 5 g of Cu is calculated as follows:

mass of Ag = (5 × 215.6 g) /63.5 g = 16.9 g

Therefore, the mass of silver precipitated will be 16.9 g.

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The electronic configuration is also used to explain the orbitals of an atom and it helps to determine the physical and chemical properties of the elements. Here the electronic configuration of chromium is 1s²2s²2p⁶3s²3p⁶4s¹3d⁵.

The distribution of electrons in the atomic orbitals is given by the electronic configuration. It is a standard notation in which all the electrons holding atomic subshells are arranged in a sequence. Each element has a unique electronic configuration.

The electronic configuration of an element can be written in two ways, in standard notation, and in condensed form. In the case of elements with larger atomic numbers, the electronic configuration becomes lengthy in standard notation. So in such cases condensed form is generally used.

In condensed form, Cr: [Ar] 3d⁵4s¹

There are six unpaired electrons in [Cr(H₂O)₆]³⁺ and it is a paramagnetic complex.

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Answer:

2.02 L

Explanation:

The volume of a gas is directly proportional to its temperature, according to the ideal gas law. This means that as the temperature of a gas decreases, its volume will also decrease.

To determine the new volume of the balloon as the temperature drops from 311 K to 295 K, we can use the ideal gas law to calculate the new volume of the gas:

PV = nRT

Where:

P = the pressure of the gas (assuming it remains constant)

V = the new volume of the gas

n = the number of moles of the gas (assuming it remains constant)

R = the universal gas constant

T = the new temperature of the gas

Since we know the pressure, number of moles, and universal gas constant, we can solve for the new volume of the gas by substituting in the known values and solving for V:

V = nRT / P

Substituting in the values from the problem, we get:

V = (n * R * 295 K) / P

Since the number of moles and the universal gas constant are constants, we can simplify the equation to:

V = k / P

Where k is a constant equal to the product of the number of moles, the universal gas constant, and the original temperature of the gas (in this case, k = n * R * 311 K).

Since we know the value of k and the pressure of the gas, we can solve for the new volume of the gas by substituting in the known values:

V = (k / P)

= (n * R * 311 K) / P

= (2.07 L * 8.31 L * atm / mol * K * 311 K) / P

= 2.02 L

Therefore, the new volume of the balloon is approximately 2.02 L.

a) NaFi is formed when 1 mol of NaOH is oriented to 1.0 L of a 0.5 M HF mixture.

b) To 1.0 L of a solution that is 0.8 M in both HF & KF-producing KF, 0.3 mol of KOH is added.

An acid generates the hydronium ions H3O+ in an aqueous solution, whereas a base creates the hydroxide ions OH. Water and salt are the byproducts of the neutralization process that occurs when an acid and a base interact.

A) 1 mol of NaOH is counted to 1.0 L of a 0.5 M HF resolution.

The reaction involved in this is:

NaOH + HF → NaF + H2O

acid: HF

base: NaOH

species that neither creates an acid and neither a base nor a salt NaF

b) To 1.0 L of a mixture that seems to be 0.8 M in both HF as well as KF, 0.3 mol of KOH is added.

The reaction involved in this is:

KOH + HF → KF + H2O

acid: HF

base: KOH

the mixture that could be neither an acid nor a base or salt delivered: KF

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The question is -

The preparations of two aqueous solutions are described below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself.

For each of the questions, write the chemical formulas of the species that will act as acids, the formulas of the species that will act as bases, and the formulas of the species that will act as neither acids nor bases. Note that HF is a weak acid.

A) 1 mol of NaOH is added to 1.0 L of a 0.5 M HF solution.

B) 0.3 mol of KOH is added to 1.0 L of a solution that is 0.8 M in both HF and KF.

The sodium is oxidized while the chlorine atom is reduced from the equation.

We know that a redox reaction is the kind of reaction that involves the change in the oxidation number of a substance. The oxidation number is the magnitude of the charge that a specie appears to have. In this case, we can see that the reaction that has been shown here has to do with a change in the oxidation number of the species that are involved in the reaction from left to right.

We can see that on the right hand side, the sodium and chlorine are both in an oxidation state of zero. If we move to the right hand side, the oxidation number of the sodium can be seen to haver increased to +1 while the oxidation n umber of the chlorine can be seen to have been decreased to - 1.

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Answer:CH

3

​

CH

2

​

CH

2

​

C≡C−CH

3

​

Explanation:

The volume of the object is 1.5ml.

This refers to the space occupied within the boundaries of an object in three-dimensional space. It is also called the capacity of the object.

In the question:

ρ = 400 g/mL

m = 600 mg

v = ?

Formular for calculating density ρ:

ρ = m/v

Where,

ρ= Density of the object

m= Mass of the object

v = volume of the object

Were are given the values of density and mass in the question. We are to calculate the volume.

Makinig v subject of the formular we have:

v = m/ρ

v =  600 mg

      400 g/mL

v = 1.5ml

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Based o the percentage masses of the elements, the empirical formula of epinephrine is C₉H₁₃O₃N.

The empirical formula of epinephrine is calculated below as follow:

Mole ratio of the elements are determined:

Carbon = 59% / 12

Carbon = 0.049 moles

Hydrogen = 7.1% / 1

Hydrogen = 0.07 moles

Oxygen = 26.2% / 16

Oxygen = 0.016

Nitrogen = 7.7% / 14

Nitrogen = 0.0055

The simplest whole-number ratio will be:

C = 0.049/0.0055

C = 9

Hydrogen = 0.07/0.0055

H = 13

Oxygen = 0.016/0.0055

Oxygen = 3

Nitrogen = 0.0055/0.0055

Nitrogen = 1

The empirical formula = C₉H₁₃O₃N

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If the reaction is exothermic as described, a rise in temperature will thus trigger the opposite reaction, which will result in a decrease in the amount of the products and an increase in the number of reactants. The reverse outcome will occur if the temperature is lowered.

B. The concentration of the products increases.

A reaction is defined as exothermic if the overall standard enthalpy change (H) is negative. Exothermic processes typically produce heat. Exergonic reaction, which the IUPAC defines as "... a reaction for which the overall standard Gibbs energy change G is negative," is frequently mistaken with the phrase.

Because "H" contributes significantly to "G," a strongly exothermic process is typically also exergonic. Exothermic and exergonic chemical reactions make up the majority of the impressive demonstrations in schools.

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the overall balanced cell reaction for the following voltaic cell

Hg2Cl2(s) + Cu(s) ® 2Hg(l) + Cu2+(aq) + 2Cl-(aq)

The cell reaction, also known as the total reaction, is the reaction that occurs within the cell as a whole and is expressed under the assumption that the right electrode is the cathode and that the spontaneous reaction is the one in which reduction takes place in the right-hand compartment.Galvanic, also known as Voltaic, and electrolytic cells are the two varieties of electrochemical cells. While electrolytic cells utilize non-spontaneous reactions and therefore need an external electron source, such as a DC battery or an AC power source, galvanic cells get their energy from spontaneous redox reactions.

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In a gas chromatogram, the retention time of compounds A and B are 1.25 min and 1.45 min, respectively. If both of these compounds have similar structures, Compound A has a higher boiling point.

Gas chromatogram is a widely used technique in analytical chemistry for the separation and investigation of compounds that may evaporate without disintegrating (GC). The purity of a substance can be assessed by GC, as can the components of a mixture. Preparative chromatography can make use of GC to separate pure compounds from a mixture. Other names for gas chromatography include vapor-phase chromatography (VPC) and gas-liquid partition chromatography (GLPC). These alternative names and the accompanying acronyms are frequently used in scientific literature. Gas chromatography is a technique for separating compounds in mixtures by injecting a gaseous or liquid sample into a mobile phase, which is usually referred to as the carrier gas, and passing the gas through a stationary phase. The mobile phase is often composed of an inert gas or an unreactive gas, such as helium, argon, nitrogen, or hydrogen. The stationary phase is a minuscule layer of viscous liquid on a surface of solid particles on an inert solid support in a piece of glass or metal tubing known as a column.

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Combustion of 31.68 g of a compound containing only carbon, hydrogen, and oxygen produces 36.67 g CO₂ and 15.01 g H₂O. the empirical formula of the compound is CH₂O.

given that :

mass of CO₂ = 36.67 g

mass of H₂O = 15.01 g

total mass of compound = 31.68 g

moles of CO₂ = 36.67 / 44

                       = 0.833 mol

moles of C = 0.833 mol

moles of  H₂O = 15.01 / 18

                        = 0.833 mol

moles of H = 2(0.833)

                  = 1.667 mol

mass of carbon = 0.833 × 12

                           = 10 g

mass of hydrogen = 1.667 × 1

                               = 1.667 g

mass of oxygen = 31.68 - ( 10 + 1.67 )

                             = 20 g

moles of Oxygen = 20 / 16

                             = 1.25 mol

dividing by the smallest one :

C = 0.833 / 0.833 = 1

H = 2

O = 1

The empirical formula is CH₂O.

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Choloracetic acid, 12 g. The colorless solution of the white crystalline substance known as chloroacetic acid.

Up to 80% of the solution can be acidic. By inhalation, ingestion, and skin contact, it is poisonous. It corrodes flesh and metals.

The equilibrium of the buffer is

CH2ClCO2 + H+ HCH2ClCO2

PKA = Log Ka

Ka: [CH2ClCO2] = 1,3 x 103. "H+" / "HCH2ClCO2"

Using the Henderson-Hasselbalch formula:

pH equals pka plus log10 [A]/[HA].

2,89 + log10 [A] / [HA] = 3,01

1,318 = [A⁻] / [HA]

Since chloroacetic acid (HA) has a molar concentration of 0,20M

[A⁻] = 0,26 mol/L

There are 500 mL plus 0.5 L.

Chloroacetic acid is equal to 0,26 mol/L x 0,5 l, or 0,13 moles. By weight:

Choloracetic acid is equal to 0,13 mol (94,5g / 1mol) = 12 g.

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Answer:

In individuals with asthma, the airways (bronchial tubes) that carry air in and out of the lungs become narrowed, swollen, and inflamed due to an overactive immune response. This can lead to difficulty breathing, wheezing, coughing, and other symptoms.

During an asthma attack, the muscles around the airways may become tight and the lining of the airways may produce excessive mucus, further narrowing the airways and making it difficult for oxygen to reach the cells in the body. This can lead to low oxygen levels in the blood and cause symptoms such as shortness of breath and chest tightness.

In addition to these physical changes in the airways, individuals with asthma may also have an overactive immune response to certain triggers, such as allergens or irritants, which can further worsen inflammation and airway narrowing.

Treatment for asthma typically involves using medications to control inflammation and relax the muscles around the airways, as well as avoiding triggers that can worsen symptoms. By properly managing their asthma, individuals with the condition can often lead normal, active lives.

Explanation:

The tubes that carry air into and out of the lungs become irritated and constricted in people with asthma. This can make it difficult for air to enter and exit the lungs, which can cause symptoms including coughing, wheezing, and breathing problems.

Being less able to enter the lungs and reach the body's cells is one-way oxygen molecules can interact differently with the systems and cells in people with asthma. It is more difficult for oxygen to enter the lungs when the airways are constricted, which can result in a reduction in the amount of oxygen reaching the body's cells. This may cause symptoms including trouble breathing and difficulty engaging in physical activity.

In people with asthma, oxygen molecules can also affect the cells and systems in different ways by inducing an immunological reaction. When allergens or pollutants are breathed by asthmatic patients, their immune systems may be overly sensitive to them and may respond inappropriately. This may result in the release of inflammatory substances, which may further constrict the airways and hinder the absorption of oxygen by the lungs.

Inflammation and immune system overactivity can generally disrupt how oxygen molecules interact with the body's processes and cells in people with asthma, which can cause breathing problems and other symptoms.

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Alkynes are hydrocarbons containing a carbon-carbon triple bond. Their general formula is CnH2n-2 for molecules with triple bonds.

Alkynes undergo many of the same reactions as alkenes but can react twice due to the presence of two p-bonds on the triple bond. Ethyne is commonly known by the trivial name acetylene. It is the simplest alkyne, consisting of two carbon atoms joined by a triple bond, where each carbon can bond to a hydrogen atom.

Molecules containing a triple bond between two carbon atoms are called alkynes. A triple bond consists of one σ bond and two π bonds. As for the hydrogen atoms, this compound is unsaturated, so the extra electrons are exchanged with two carbon atoms forming a double bond.

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An organism with the genotype of AaXx can produce gametes containing  either AX, Ax, aX, ax  if the two genes are unlinked .

An organism's genotype is the complete set of its genetic material. [1] Genotype can also be used to refer to the alleles or variants that an individual has at a particular gene or locus. The number of alleles an individual can have for a particular gene depends on the number of copies of each chromosome (also called polyploidy) found in that species. Diploid species like humans have two complete sets of chromosomes. This means that each individual has her two alleles for a particular gene. If both alleles are the same, the genotype is said to be homozygous. If the alleles are different, the genotype is called heterozygous.

Genotypes contribute to the phenotype, observable traits and characteristics of an individual or organism. The extent to which genotype influences phenotype depends on the trait.

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Answer:Co2O3

Explanation:

look it up

Answer:

Co2O3

Explanation:

when you write any formula you apply the criss-cross method where you write the charges of each substance

Co3 and O2

applying the criss-cross method( you switch the charges number, in this case you switch the 2 and the 3) :-

Co2 and O3

so Co2O3

AlCl 3 is Lewis base,PH3 is Lewis base,BH3 is Lewis acid ,BF3 is Lewis acid.

The crucial step is AlCl 3 accepting a chloride ion lone-pair to generate AlCl 4 and the highly acidic, or electrophilic, carbonium ion. Lewis bases:  RCl +AlCl 3 → R + + AlCl 4−, etc. [ edit] The highest occupied molecular orbital (HOMO) of an atomic or molecular species that is strongly localised is known as a Lewis base.Lewis acid borane (BH3) has three hydrogen atoms and one boron atom in its molecule. The Lewis structure of borane features a single connection connecting each hydrogen atom to boron (BH3). Only three bonds surround the boron atom, and there are no lone pairs on the atom itself. We will learn how to draw the BH3 Lewis structure in this lesson.Only six electrons orbit the boron atom, according the Lewis structure of BF3. As a result, the boron atom's octal is incomplete. Borane BF3 is therefore regarded as a Lewis acid.A highly poisonous gaseous chemical is phosphorus. Three sigma bonds and one lone pair are present around the phosphorus atom in the Lewis structure of phosphine (PH3). Hydrogen and phosphorous atoms have no charges. The trigonal pyramidal shape of PH3 is a fundamental shape.

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Specific heat capacity of a substance is the amount of heat required to raise the temperature by one degree Celsius of one gram of a substance. Therefore, the total enthalpy of the given reaction is -781.44KJ.

Enthalpy term is basically used in thermodynamics to show the overall energy that a matter have. Mathematically, Enthalpy is directly proportional to specific heat capacity of a substances.

Mathematically,

q = n ×ΔH

where

q = amount of heat

n = no of moles

ΔH = enthalpy

n = w / M.M

w = given mass

M.M = molar mass

4CH[tex]_3[/tex]NH[tex]_2[/tex]+2H[tex]_2[/tex]O[tex]\rightarrow[/tex]3CH[tex]_4[/tex]+CO[tex]_2[/tex]+4NH[tex]_3[/tex]   ΔH=-138.8KJ

number of moles of methyl amine=175/31.06

                                                       =5.63moles

enthalpy for reaction of one mole of methyl amine=-138.8KJ

enthalpy for reaction of 5.63 mole of methyl amine=-138.8KJ×5.63

                                                                                     =-781.44KJ

Therefore, the total enthalpy of the given reaction is -781.44KJ.

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Which type of arrangement of atoms could a molecule with five covalent bond (and any percentage of lone pairs) have? Trigonal bipyramidal, square pyramidal

A cell is the smallest unit of a material that retains its composition and properties. It is made up of 2 or more atoms that are joined together by covalent bond. Chemistry is built on molecules. The element sign and a suffix indicating the number of ions are used to identify molecules. A molecule is the smallest unit of any material that is composed of one or more elements and is capable of existing independently while maintaining all of the substance's physical and chemical properties. Further atom division occurs within molecules. For instance, the symbols for the oxygen atom and molecule are O and O2, respectively.

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